From the top of a tower, of 100 m height, the angles of depression of two objects, 200 3 m apart on the horizontal plane on a line passing through the foot of the tower and on the same side of the tower are 45 o - A and 45 o + A . Then angle A is equal to

Step-by-step solution

Let O P be tower and objects are placed at A & B , then The distance between the objects = 100 c o t 45 o - A - c o t 45 o + A from Δ O P A & Δ O P B = 100 1 + t a n A 1 - t a n A - 1 - t a n A 1 + t a n A = 100 1 + tan ⁡ A 2 - 1 - tan ⁡ A 2 1 - t a n 2 A = 100 . 4 tan ⁡ A 1 - t a n 2 A = 200 tan ⁡ 2 A ∴ 200 3 = 200 tan ⁡ 2 A ⇒ tan ⁡ 2 A = 1 3 ∴ 2 A = 30 o ⇒ A = 15 o